Answer:
[tex]\sf \angle A = 53^\circ [/tex]
Step-by-step explanation:
Given:
- [tex] \sf \overline{AB} [/tex] is diameter.
- [tex]\sf \angle B = 37^\circ [/tex]
To find:
- [tex]\sf \angle A = ? [/tex]
Solution:
Recognize that for any inscribed angle in a circle, where the angle is formed by a diameter and a chord, the inscribed angle is always a right angle (i.e., [tex]\bold{\sf 90^\circ}[/tex]).
Since [tex]\bold{\sf \overline{AB}}[/tex] is a diameter of circle [tex]\bold{\sf O}[/tex], [tex]\bold{\sf \angle C}[/tex] (which subtends this diameter) must indeed be [tex]\bold{\sf 90^\circ}[/tex].
Apply the property that the sum of angles in a triangle is [tex]\bold{\sf 180^\circ}[/tex].
In triangle [tex]\bold{\sf ABC}[/tex]
We have
[tex]\sf \angle A + \angle B + \angle C = 180^\circ [/tex]
Substitute [tex]\bold{\sf \angle C = 90^\circ}[/tex] into the equation:
[tex]\sf \angle A + 37^\circ + 90^\circ = 180^\circ [/tex]
[tex]\sf \angle A + 127^\circ = 180^\circ [/tex]
[tex]\sf \angle A = 180^\circ - 127^\circ [/tex]
[tex]\sf \angle A = 53^\circ [/tex]
Therefore, the measure of [tex]\bold{\sf \angle A}[/tex] in triangle [tex]\bold{\sf ABC}[/tex] is [tex]\bold{\sf \boxed{53^\circ} }[/tex].
